If X and Z are independent and P(X) = .6, P(Z) = .3, compute:
A) P(x U z)
B) P (x intersection z)

Question

If X and Z are independent and P(X) = .6,   P(Z) = .3, compute:
A)  P(x U z)
B)  P (x intersection z)

anonymous · Answer

A)0.6+0.3=0.9
b)0.6*0.9=0.54

anonymous · Answer

the upside down U means intersection right

anonymous · Answer

Nope U means Union

anonymous · Answer

i know but i couldnt do the upside down one on the computer so i put intersection

anonymous · Answer

The probability of a person over the age of 53 developing cancer is .06, and the probability that she/he develops heart problems is .08.   If a patient over 53 is selected at random, what is the probability that the patient will develop:

A)   None of these deseases? 
B)   At least one of the diseases? 
C)   Both of the diseases?

anonymous · Answer

can u help me on this one?

anonymous · Answer

please

zarkon · Answer

$$P(X\cup Z)=.72$$

$$P(X\cap Z)=.18$$